Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
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 *                      @@@.                                                N@BML2U2UUU12F15252525251515Jk@@B
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 *                     @@@7                                                   B@B@5Yj12F152F1F1F25252515jFB@B
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 *            B@B ,@B1          L0EZZG0F7:                                            .:,                        uB@MrP@M7
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 *            7@,      kB@U    ;r        @@@.       .@Z                GBuL@iBBi           vB@B@q BP:5@7 @u,.  @B
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 *                          .ll                                           rB.                     :
 */
/*
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██    ██  ████  ██████  ██    ******************    ████  ██            ██  ██
██    ██████  ██  ██████      * Scan this QR   *██████    ██  ██  ██  ████████
████  ████  ██  ██████      ██* code to join   *██  ████    ██  ████████  ████
██████      ████████  ██      * in the Three   *██  ████          ████  ██  ██
████              ████  ████  * Body ETO.      *  ████        ██  ████  ██  ██
██████  ██    ████████        *  --ETO manager *██  ██                ██    ██
████    ██  ██  ██    ██████  *    Ye Wenjie   *████  ████  ██  ████  ██  ████
██  ████  ██████    ████  ██  *ETO(C),1988~2018*      ██  ██  ██  ██    ██████
██  ██████  ██  ████    ██  ██******************          ██████    ██████  ██
██  ██  ██    ████    ██        ██  ██        ████████      ████  ██  ██    ██
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██  ██      ██  ████  ████  ████  ██████  ██████████████  ██████████    ██  ██
██  ██      ██  ██    ██████  ██  ██  ██████████  ██  ██      ████    ██    ██
██  ██████████  ██████  ████            ██    ████████  ██  ████    ██████████
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██████████████████████████████████████████████████████████████████████████████
    扫    描    二    维    码    加    入    三    体    E    T    O
*/
/*****************
 *   I<3   C++   *
 *  IWANT ANYAC  *
 * ILOVEPROGRAM! *
 *IT'SINTERESTING*
 * ILOVEPROGRAM! *
 *  INCONTESTS   *
 *   GETSCORE    *
 *    ACCODE     *
 *     LETS      *
 *      GO       *
 *Written by:    *
 *    charlieyan *
 *****************/
#include <assert.h>
#include <ctype.h>
#include <errno.h>
#include <float.h>
#include <fstream>
#include <iomanip>
#include <iostream>
#include <limits>
#include <locale>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <wchar.h>
#include <wctype.h>
#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <numeric>
#include <list>
#define LL long long
#define pb push_back
#define pob pop_back
#define mp make_pair
#define FILL(a,v) memset((void*)a,v,sizeof(a))
#define FILL0(a) FILL(a,0)
#define FILL1(a) FILL(a,-1)
#define FILLBIG(a) FILL(a,0x3f)
#define ALL(a) (a).begin(),(a).end()
#define Inf ((int)0x3f3f3f3f)
#define INF ((long long)0x3f3f3f3f3f3f3f3f)
#define Mod 1000000007
#define foreach(it,x) for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)
#define FOR(i,x) for(int i=0;i<x;i++)
#define RFOR(i,x) for(int i=x-1;i>=0;i--)
#define FFOR(i,a,b) for(int i=a;i<b;i++)
#define FRFOR(i,a,b) for(int i=a-1;i>=b;i--)
#define SZ(a) ((int)(a).size())
#define FI first
#define SE second
#define OVER(...) return printf(__VA_ARGS__),0
#define fastio ios::sync_with_stdio(0)
#define filei(f) freopen(f,"r",stdin)
#define fileo(f) freopen(f,"w",stdout)
#define eprintf(...) fprintf(stderr,__VA_ARGS__)
#define MASK(a) (1<<(a))
#define MASKX(x,a) ((x)<<(a))
#define BIT(a,b) (((a)>>(b))&1)
#define cddx(a) priority_queue<a >
#define cxdd(a) priority_queue<a,vector<a >,greater<a > >
#define deval(a) {cerr<<#a<<" = "<<a<<endl;}
#define dearr(be,en) {cerr<<#be;cerr<<" = {";for(auto i=be;i!=en;i++)cerr<<' '<<*i;cerr<<" }\n";}
#define reg register
#define prtime eprintf("\n\nTime used: %dms\n",clock());
#define PI 3.1415926535897932384626
#define TODEG(a) (a*180/PI)
#define y0 ykangjimutigeidiqiuyipianjingtu
#define y1 yxiaomiexuduhuanrenleiwannianankang
#define endl '\n'
#define left leftxiaomiexudurenrenyouze
#define hash hashxushibingduyouweilifushiyiqiedewuti
//抗击母体，给地球一片净土
//消灭徐毒，还人类万年安康
using namespace std;
typedef pair<int,int> pii;
typedef unsigned long long UINT;
typedef unsigned int uint;
typedef long long INT;
typedef long double ldouble;
typedef pair<INT,INT> PII;
typedef stringstream sst;
typedef istringstream iss;
typedef vector<int> vint;
typedef vector<INT> VINT;
typedef vector<pii> vpii;
typedef vector<PII> VPII;
typedef priority_queue<int> pqi;
typedef priority_queue<INT> PQI;
typedef priority_queue<pii> pqp;
typedef priority_queue<PII> PQP;
typedef priority_queue<int,vint,greater<int> > xqi;
typedef priority_queue<INT,VINT,greater<INT> > XQI;
typedef priority_queue<pii,vpii,greater<pii> > xqp;
typedef priority_queue<PII,VPII,greater<PII> > XQP;
struct fastIO
{
	inline fastIO operator>>(int& num)
	{
		num=0;
		char c=getchar();
		while(c!='-'&&(c<'0'||c>'9'))c=getchar();
		bool foo=0;
		if(c=='-')
		{
			foo=1;
			c=getchar();
		}
		while(c>='0'&&c<='9')
		{
			num=(num<<3)+(num<<1)+c-'0';
			c=getchar();
		}
		if(foo)num=-num;
		return *this;
	}
	inline fastIO operator>>(INT& num)
	{
		num=0;
		char c=getchar();
		while(c!='-'&&(c<'0'||c>'9'))c=getchar();
		bool foo=0;
		if(c=='-')
		{
			foo=1;
			c=getchar();
		}
		while(c>='0'&&c<='9')
		{
			num=(num<<3)+(num<<1)+c-'0';
			c=getchar();
		}
		if(foo)num=-num;
		return *this;
	}
	inline fastIO operator>>(float& num)
	{
		scanf("%f",&num);
		return *this;
	}
	inline fastIO operator>>(double& num)
	{
		scanf("%lf",&num);
		return *this;
	}
	inline fastIO operator>>(long double& num)
	{
		scanf("%Lf",&num);
		return *this;
	}
	inline fastIO operator>>(char& num)
	{
		num=getchar();
		while(num==' '||num=='\n')num=getchar();
		return *this;
	}
	inline fastIO operator>>(char* num)
	{
		int cnt=0;
		char c=getchar();
		while(c==' '||c=='\n')c=getchar();
		while(c!=' '&&c!='\n')
		{
			num[cnt++]=c;
			c=getchar();
		}
		num[cnt]=0;
		return *this;
	}
	inline fastIO operator>>(string& num)
	{
		cin>>num;
		return *this;
	}
	inline void printInt(const int& num)
	{
		if(num<10)
		{
			putchar(num+'0');
			return;
		}
		printInt(num/10);
		putchar((num%10)+'0');
	}
	inline void printINT(const INT& num)
	{
		if(num<10)
		{
			putchar(num+'0');
			return;
		}
		printINT(num/10);
		putchar((num%10)+'0');
	}
	inline fastIO operator<<(const int& num)
	{
		if(num<0)putchar('-'),printInt(-num);
		else printInt(num);
		return *this;
	}
	inline fastIO operator<<(const INT& num)
	{
		if(num<0)putchar('-'),printINT(-num);
		else printINT(num);
		return *this;
	}
	inline fastIO operator<<(const float& num)
	{
		printf("%.10f",num);
		return *this;
	}
	inline fastIO operator<<(const double& num)
	{
		printf("%.16lf",num);
		return *this;
	}
	inline fastIO operator<<(const long double& num)
	{
		printf("%.20Lf",num);
		return *this;
	}
	inline fastIO operator<<(const char& num)
	{
		putchar(num);
		return *this;
	}
	inline fastIO operator<<(const char* num)
	{
		for(int i=0;num[i];i++)
			putchar(num[i]);
		return *this;
	}
	inline fastIO operator<<(const string& num)
	{
		cout<<num;
		return *this;
	}
}fio;
//Coded by Charlieyan
//版权所有，翻印必究

struct Segtree{
	int N;
	vint val;
	Segtree(){}
	Segtree(int n){
		N=1;
		while(N<n){
			N<<=1;
		}
		val.assign(N*2,Inf);
	}
	int getval(int k){
		k+=N;
		int ans=Inf;
		while(k){
			ans=min(ans,val[k]);
			k>>=1;
		}
		return ans;
	}
	void change(int a,int b,int num){
		a+=N;
		b+=N;
		while(a<b){
			if(a&1){
				val[a]=min(val[a],num);
				a++;
			}
			if(b&1){
				b--;
				val[b]=min(val[b],num);
			}
			a>>=1;
			b>>=1;
		}
	}
};

int n;
INT k;
INT a[100005],b[100005];
INT f[100005];
VINT xs;
Segtree seg(200005);

int getid(INT x){
	return lower_bound(ALL(xs),x)-xs.begin();
}

INT solve(VPII &vp){
	int N=SZ(vp);
	if(!N){
		return 0;
	}
	xs.pb(0);
	xs.pb(k);
	for(int i=0;i<N;i++){
		xs.pb(vp[i].first);
		xs.pb(vp[i].second);
	}
	sort(ALL(xs));
	xs.resize(unique(ALL(xs))-xs.begin());
	int X=SZ(xs);
//	cerr<<X<<endl;
	Segtree seg(X);
	VINT dpl(N,INF),dpr(N,INF);
//	cerr<<13<<endl;
	for(int i=N-1;i>=0;i--){
//		cerr<<i<<endl;
		int l=getid(vp[i].first),r=getid(vp[i].second);
		int j=seg.getval(l);
		if(j==Inf){
			dpl[i]=0;
		}else{
			INT dis=vp[j].first-vp[i].first;
			if(dis<0){
				dis+=k;
			}
			dpl[i]=dpl[j]+dis;
		}
		j=seg.getval(r);
		if(j==Inf){
			dpr[i]=0;
		}else{
			INT dis=vp[j].first-vp[i].second;
			if(dis<0){
				dis+=k;
			}
			dpr[i]=dpl[j]+dis;
		}
		if(l<=r){
			seg.change(0,l,i);
			seg.change(r+1,X,i);
		}else{
			seg.change(r+1,l,i);
		}
//		cerr<<i<<endl;
	}
	INT ans=INF;
//	cerr<<ans<<endl;
	for(int i=0;i<N;i++){
//		cerr<<i<<' '<<N<<endl;
		int l=getid(vp[i].first),r=getid(vp[i].second);
		int j=seg.getval(l);
		if(j>i){
			ans=min(ans,dpl[i]);
		}
		j=seg.getval(r);
		if(j>i){
			ans=min(ans,dpr[i]);
		}
//		cerr<<i<<' '<<N<<endl;
	}
//	cerr<<13<<endl;
	return ans;
}

int main(){
	fio>>n>>k;
	for(int i=0;i<n;i++){
		fio>>a[i];
		int b;
		fio>>b;
		f[i]=(b==2);
		if(b==1&&a[i]*2>k){
			OVER("-1");
		}
	}
	for(int i=0;i<n;i++){
		b[i+1]=b[i]+a[i]*2;
	}
	VPII vp;
	for(int i=0;i<n;i++){
		if(!f[i]){
			INT l=((-b[i])%k+k)%k;
			INT r=((-b[i+1])%k+k)%k;
			vp.pb(mp(l,r));
		}
	}
	INT faq=solve(vp);
	//cerr<<faq<<endl;
	fio<<b[n]+faq<<endl;
	return 0;
}